# hard parametrization problem

Find a parametrization of $\vec{r}(x,y,z)$ in terms of $t$ such that,

$\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2 + \left(\dfrac{dz}{dt}\right)^2 = 4t^2$

$xy = t$

How can I find such $\vec{r}(x,y,z)$ in terms of $t$? I find it hard to do so.

-

There should be solutions for many choices of $x(t)$ and $y(t)$. Start by parametrizing any point $\vec p$ in the plane that has a velocity no greater than $2t$ and manages to be somewhere on the hyperbola $xy=t$ at time $t$. Then $$\left({dx\over dt}\right)^2 + \left({dy\over dt}\right)^2 + \left({dz\over dt}\right)^2 = (\textrm{velocity of } \vec p)^2 +\left({dz\over dt}\right)^2{\textrm.}$$
Then it will be possible to choose $z(t)$ so that $\left({dz\over dt}\right)^2 = 4t^2 - (\textrm{velocity of } \vec p)^2$, and that will give the desired result.
A simple pair of functions to try is $x(t)=t$ and $y(t)=1$. Then $\left({dx\over dt}\right)^2 + \left({dy\over dt}\right)^2 = 1$. Therefore you want to choose $z(t)$ so that $\left({dz\over dt}\right)^2 = 4t^2-1$. A little calculus gives $$z(t)=\frac{1}{2} t \sqrt{-1+4 t^2}-\frac{1}{4} \text{Log}\left[2 t+\sqrt{-1+4 t^2}\right]{\textrm.}$$